Arc-length change of variable in the inviscid Burgers equation.

51 Views Asked by At

Being quite new to the world of PDEs, I would like your help regarding a specific change of variable. Namely, I consider the inviscid Burgers equation : \begin{equation} u_t+uu_x=0 \end{equation} And I would like to do the change of variable $(t,x) \mapsto (t,s(t,x))$ where : \begin{equation} \quad ds^2=dx^2+du^2 \quad \text{or} \quad ds = \sqrt{1+u_x^2}dx \end{equation} Using the chain rule, I have : \begin{equation} \begin{cases} u_t = \frac{du}{dt} + \frac{du}{ds} \frac{ds}{dt} \\ u_x = \frac{du}{dt} \frac{dt}{dx} + \frac{du}{ds}\frac{ds}{dx} = u_s \sqrt{1+u_x^2} = \frac{u_s}{\sqrt{1-u_s^2}} \end{cases} \end{equation} If I then replace into the PDE, I obtain : \begin{equation} u_t + u_s s_t + \frac{[u^2]_s}{2 \sqrt{1-u_s^2}} = 0 \end{equation} This is where I request your help. I would like to know if the computations above make sense (provided that $u_s^2$ is never equals to 1), and how can I handle the term $s_t$ since I do not know its explicit expression ?

Moreover, if I want to do the reverse change of variable to get $u(t,x)$, how do I proceed ?